Correct optocoupler output emitter configuration +ve ground circuit

Hi.

Referring to the circuit below, I’m not sure the optocoupler configuration is correct? It is in linear mode and there is no resistor to modify the output.

I wish to confirm, that as far as the optocoupler transistor is concerned, that it is properly set up as a switch to emulate the pushbutton. I’m fairly sure it is…

EDIT: it’s likely not… see my second post.

Input 5V. Could be Arduino digital pin high/low.

Output is connected to -5V (-VE polarity) motor control logic.

In this case the emitter is connected to -5V and the load, while the collector is connected to GND (+VE).

Note: The motor symbol is generic and implies control circuitry before the motor. The motor may be activated manually with a push button. The 1M pull down resistor appears to prevent feedback during manual operation. Without it the optocoupler output generates feedback through the +VE GND line. At least that’s what it indicates in EveryCircuit.

6fb6ae3c9f54af22c86e8a39af61a61f6c076b06.jpg

Thanks. The Load should perhaps be in series with the collector and ground.

EDIT: There are two possibilities - emitter to +VE ground and load to collector produces the highest current driving an LED.

There are two possibilities - emitter to +VE ground and load to collector produces the highest current driving an LED.

I set up a breadboard with Arduino input on one side and -5v power on the other and a Vishay ILCT6 bridging the gap. Not as complex as I first imagined. Swapping the hook up wires between collector and emitter very quickly established the highest current based on LED brightness. Had to think hard about hooking it all up with negative polarity.

An optocoupler with a high CTR should make it a functioning project, emulating the pushbutton quite well.

first of all, unless I have the wrong datasheet, this is an NPN output transistor. The emitter goes directly to ( -). The load goes on the (+) side.

next, what are you doing with it. You can only drawn 150 mW load Max. Not enough for most motors. You need a second transistor to handle the load.

what type Load do you have, and how much current does it draw.

Hi,
I think what you need it is something like the attached drawing circuit. You can use any mosfet that can handled the load of 12 volts .6 amps or at least 1 amp.

arduino-fan-cicuit (1).pdf (20.2 KB)

Thanks. The pushbutton activates TTL motor control logic. I wish to emulate/replace use of the pushbutton with the optocoupled input/output. The motor control logic is +ve ground, hence the optocoupler.

Motors are -12V behind the control circuit, which is activated by the pushbutton

I am substituting a mechanical switch with the NPN transistor of the optocoupler. It appears that the emitter goes to +ve ground and the collector goes to the load (-5V) and then to ground in series. My second schematic..

There are two ways of doing this. Emitter to +ve GND or collector to +ve GND. I will check again, but Emitter to +ve GND appears to be correct, given the polarity of the motor control circuit. I've had more time to play with it.

Hi,

Attached it is a circuit that will allow you to use the optical coupler as a switch. Red box are a simulation because those components are not in the simulation list. So i have to use a simulation to use it for how to wire the circuit. Hope this is what are you looking for. The Q2 transistor need to be selected that will able to drive the fan load.

optical-fan-control.pdf (21.4 KB)

Is there a requirement that the fan circuit have the + side grounded?

Thank you, I appreciate the efforts of all...

The equipment is 20yo and has a +ve ground circuit. I am interfacing with this equipment to automate its functions.

There is no fan or motor power to consider. It's just switching at +5 and -5V (TTL) each side of the optocoupler.

Since beginning this thread I have been able to breadboard the setup and find that the highest current flow is with the emitter to +ve ground and the collector to the load.

The circuit simulators I am using do not show a difference in current flow with the emitter and collector connections swapped. This was confusing.

I am satisfied that the emitter goes to ground and the collector to load, though the polarities are reversed. This would seem to be correct, though I was unsure to begin with...